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(D) Let the thermal resistance of a rod with conductivity $K$ be $R = \frac{l}{Ka}$.Then,the resistance of a rod with conductivity $2K$ is $R' = \frac

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$\frac{l}{2ka}$

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(D) Let the thermal resistance of a rod with conductivity $K$ be $R = \frac{l}{Ka}$.Then,the resistance of a rod with conductivity $2K$ is $R' = \frac{l}{(2K)a} = \frac{R}{2}$.From the equivalent circuit diagram,the upper branch consists of three parts in series:$1$. Two rods of $2K$ in parallel: $R_1 = \frac{(R/2)(R/2)}{(R/2) + (R/2)} = \frac{R/4}{1} = \frac{R}{4}$.$2$. Two rods of $K$ in parallel: $R_2 = \frac{R \cdot R}{R + R} = \frac{R}{2}$.$3$. Two rods of $2K$ in parallel: $R_3 = \frac{(R/2)(R/2)}{(R/2) + (R/2)} = \frac{R}{4}$.The total resistance of the upper branch is $R_{upper} = R_1 + R_2 + R_3 = \frac{R}{4} + \frac{R}{2} + \frac{R}{4} = R$.The lower branch is a single rod of conductivity $K$,so $R_{lower} = R$.The effective resistance $R_{eq}$ between $A$ and $B$ is the parallel combination of the upper and lower branches:$R_{eq} = \frac{R_{upper} \cdot R_{lower}}{R_{upper} + R_{lower}} = \frac{R \cdot R}{R + R} = \frac{R}{2}$.Substituting $R = \frac{l}{Ka}$,we get $R_{eq} = \frac{1}{2} \left( \frac{l}{Ka} \right) = \frac{l}{2Ka}$.

Find the effective thermal resistance between $A$ and $B$ of a cube made up of $12$ rods of the same dimensions,with the given thermal conductivities as shown in the figure. [ $l =$ length of rod,$a =$ cross-sectional area of rod ]

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